Question

the general social survey polled a sample of 209 people aged 18-30 in the year 2000,asking...

the general social survey polled a sample of 209 people aged 18-30 in the year 2000,asking them how many hours per week they spent on the internet. the sample mean was 6.75 with a standard deviation of 7.7. a second sample, the mean was 7.34 with a standard deviation 10.9.assume these are simple random sample from population of people aged 18-30.can you conclude that the mean number of hours per week spent on the internet increased between 2000 and 2006?. Calculate the Margin of Error for 95% confidence level.

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
In a random sample of 23 people aged 18-24, the mean amount of time spent using...
In a random sample of 23 people aged 18-24, the mean amount of time spent using the internet is 45.1 hours per month with a standard deviation of 28.4 hours per month. Assume that the amount of time spent using the internet in this age group is normally distributed. 1. if one were to calculate a confidence interval for the population mean, would it be a z or t confidence interval? use the flow chart to explain your answer. 2....
The 2006 General Social Survey contains information on the number of hours worked by a respondent...
The 2006 General Social Survey contains information on the number of hours worked by a respondent each week. The mean number of hours worked per week is 39.04, with a standard deviation of 11.51. The sample size is 83. c. Say we increased the sample size to 10,000 and found the same mean and standard deviation. What would be the standard error? d.If we increased the sample size to 10,000 what would be the 95% confidence interval for the mean...
The 2012 general Social Survey asked a large number of people how much time they spent...
The 2012 general Social Survey asked a large number of people how much time they spent watching TVeach day. The mean number of hours was 3.09 with a standard deviation of 2.83. Assume that in a sampleof 45 teenagers, the sample standard deviation of daily TV time is 4.1 hours, and that the population ofTV watching times is normally distributed. Can you conclude that the population standard deviation ofTV watching times for teenagers differs from 2.83? Use theα= 0.10 level...
Watching TV: The General Social Survey asked a sample of 1298 people how much time they...
Watching TV: The General Social Survey asked a sample of 1298 people how much time they spent watching TV each day. The mean number of hours was 3.09 with a standard deviation of 2.87. A sociologist claims that people watch a mean of 3 hours of TV per day. Do the data provide sufficient evidence to disprove the claim? Use the =0.01 level of significance. Solve without use of TI-84 special functions Show P- Value figure Reject or Do not...
Watching TV: In 2012, the General Social Survey asked a sample of 1328 people how much...
Watching TV: In 2012, the General Social Survey asked a sample of 1328 people how much time they spent watching TV each day. The mean number of hours was 3.03 with a standard deviation of 2.68. A sociologist claims that people watch a mean of 3 hours of TV per day. Do the data provide sufficient evidence to conclude that the mean hours of TV watched per day is greater than the claim? Use the =α0.10 level of significance and...
1. (Hypothetical) The General Social Survey measures the number of hours that individuals spend on the...
1. (Hypothetical) The General Social Survey measures the number of hours that individuals spend on the internet each week. Males use the Internet 12.1 hours per week. (standard deviation 10.8; N = 264), while women use the Internet 9.10 hours per week (standard deviation 12.50; N = 300). a) Test the research hypothesis that men use the internet more hours a than women. Set alpha at .05. b) Would your decision have been different if alpha were set at .01?
The General Social Survey (GSS) is a survey of approximately 2000 adults conducted each year since...
The General Social Survey (GSS) is a survey of approximately 2000 adults conducted each year since 1972, for a total of 38,000 participants. During several years of the GSS, participants were asked how many close friends they have. The mean for this variable is 7.44 friends, with a standard deviation of 10.98. The median is 5.00, and the mode is 4.00. 4. Now pretend you randomly selected a sample of 80 people from this population. Using symbolic notation, calculate the...
1. the data collected from the 2012 General Social Survey (GSS) in which 1,298 respondents reported...
1. the data collected from the 2012 General Social Survey (GSS) in which 1,298 respondents reported on the number of hours they watched tv per day. The results show that the average (mean) number of hours spent watching tv per day is 3.089 hours with a standard deviation of .07 hours. a. Calculate 95% confidence interval to estimate the mean number of hours of tv the average American watches per day based on these data. b. What do you notice...
1. The General Society Survey asked a sample of 1200 people how much time they spent...
1. The General Society Survey asked a sample of 1200 people how much time they spent watching TV each day. The mean number of hours was 3.0 with a standard deviation of 2.87. A sociologist claims that people watch a mean of 4 hours of TV per day. Do the data provide sufficient evidence to disprove the claim? Use ? = .05 to test the hypothesis. 2. We have developed a regression equation to predict son’s height from father’s height:...
I asked a random sample of 15 adults, aged 25 - 50, how many hours per...
I asked a random sample of 15 adults, aged 25 - 50, how many hours per week they spent playing video games. The responses were as follows: 7, 15, 3, 1, 2, 2, 0, 0, 10, 5, 1, 1, 2, 0, 2 With a mean of 3.4 and a standard deviation of 4.256, determine a 90% confidence interval.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT